# 24 - Amazon Top Interview Questions

### Problem Statement :

```You are given a list of four integers nums, each between 0 and 9, in a fixed order. By placing the operators +, -, *, and / (where / is integer division) between the numbers, and grouping them with parentheses, determine whether it is possible to get the value 24.

Constraints

n = 4 where n is the length of nums

0 ≤ nums[i] ≤ 9

Example 1

Input

nums = [5, 2, 7, 8]

Output

True

Explanation

We can do (5 * 2 - 7) * 8 = 24

Example 2

Input

nums = [7, 9, 7, 4]

Output

True

Explanation

We can make 24 with (7 - (9 / 7)) * 4. Note that / is integer division so the result of 9 / 7 is floored.```

### Solution :

```                        ```Solution in C++ :

vector<int> count(vector<int>& nums, int l, int r) {
if (l == r) return {nums[l]};
vector<int> ret;
for (int j = l; j < r; j++) {
vector<int> left = count(nums, l, j);
vector<int> right = count(nums, j + 1, r);
for (auto& n : left) {
for (auto& p : right) {
ret.push_back(n + p);
ret.push_back(n * p);
ret.push_back(n - p);
if (p != 0) ret.push_back(n / p);
}
}
}
return ret;
}
bool solve(vector<int>& nums) {
int l = 0, r = 3;
auto dos = count(nums, l, r);
int ret = 0;
for (auto& n : dos) ret += n == 24;
return ret > 0;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
char[] ops = {'+', '-', '*', '/'};

public boolean solve(int[] nums) {
return check(nums);
}

private boolean check(int[] arr) {
if (arr.length == 1) {
return arr[0] == 24;
}

for (int i = 0; i < arr.length - 1; i++) {
for (char op : ops) {
int[] res = operate(arr, i, op);
if (check(res)) {
return true;
}
}
}

return false;
}

private int[] operate(int[] orig, int idx, char op) {
int[] res = new int[orig.length - 1];
for (int i = 0; i < idx; i++) {
res[i] = orig[i];
}

if (op == '+') {
res[idx] = orig[idx] + orig[idx + 1];
} else if (op == '-') {
res[idx] = orig[idx] - orig[idx + 1];
} else if (op == '*') {
res[idx] = orig[idx] * orig[idx + 1];
} else {
res[idx] = orig[idx + 1] == 0 ? 0 : orig[idx] / orig[idx + 1];
}

for (int i = idx + 2; i < orig.length; i++) {
res[i - 1] = orig[i];
}

return res;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums):
self.res = False

def go(i, ops, postfix):
# no. of ops. used is never more than
# index
if i == 4 and ops == 3:
if self.postorder_eval(postfix) == 24:
self.res = True
if ops < i - 1:
for op in "+-/*":
postfix.append(op)
go(i, ops + 1, postfix)
postfix.pop()

if i < len(nums):
postfix.append(nums[i])
go(i + 1, ops, postfix)
postfix.pop()

return

go(0, 0, [])
return self.res

# postfix is easy to evaluate and no brackets ;)
def postorder_eval(self, postfix):
res = []

for c in postfix:
if isinstance(c, str):
b = int(res.pop())
a = int(res.pop())
if c == "+":
c = a + b
elif c == "-":
c = a - b
elif c == "*":
c = a * b
elif c == "/":
if b == 0:
c = 0
else:
c = a // b
res.append(c)
else:
res.append(c)

return res[-1]```
```

## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =